Re: 2nd law of thermodynamics in question

Richard Herring wrote:
In message <1163518186.732439.209950@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul
<softwarelabus@xxxxxxxxx> writes
Richard Herring wrote:
In message <1163444775.906688.99910@xxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul
<softwarelabus@xxxxxxxxx> writes
Richard Herring wrote:
In message <1163436256.228189.76090@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul
<softwarelabus@xxxxxxxxx> writes
jambaugh wrote:
Paul wrote:

Which experiment generates more radiation? Answer: Experiment #1
[previously corrected!]

That's very close to falsifying the record. The text I quoted looked
like this:
===================================================
Which experiment generates more radiation? Answer: Experiment #2
===================================================


How dare you. I clearly placed "[previously corrected!]"




Sure about that? ;-)

Yes, I am sure about that. Experiment #1. Did you see my typo
correction within minutes after posting? Perhaps you want to be a j---.

No, it should be obvious that when I posted I had not seen your
correction.

It is courteous to view a person's entire picture / post before
commenting or criticizing.

<sigh>

I _had_ read your entire post. I'm not a mind-reader, so I don't see how
you could expect me to know that you were about to post a correction.

Your post came _after_ both my posts. Again, perhaps you would make a
great politician, especially in a smear campaign.




Doesn't take a genius, Richard, to understand experiment B radiates
more power.

And where did I suggest otherwise?

Doesn't take a genius, Paul, to understand my experiment B has a greater
power flux than A.

Experiment B will be cooler than experiment A.

Until it reaches thermal equilibrium.

Experiment B tends towards equilibrium faster than experiment A. So?

Very simple
stuff Richard. That is direct proof the 2nd Law of Thermodynamics is
incorrect.

And that's what we call a non sequitur. Since, as you pointed out, there
are many formulations of the 2nd Law, perhaps you should clarify matters
by posting the one you are using, and then explain exactly which part of
it you think is contradicted.

Then you agree such an experiment contradicts some interpretations of
the 2nd law, right?

Not any of the correct ones.

Just as I thought, which is clearly why I require predictions from QM
physicists prior to performing my sub-photon experiment. A clear and
precise prediction which you were obviously afraid to make and commit
to, as evident in our other thread. My present mistake was in not
nailing you down before hand. Now you are attempting to weasel your way
out by insinuating a thermodynamic isolated system even prohibits
gravity and magnetism from escaping. Later on in this thread I will
comment how ridiculous and useless such a law would be.



Did you see my original quote from WikiPedia.com
where they stated the following?

Quote from WikiPedia,com:
"Since any thermodynamic engine requires such a temperature difference,
it follows that no useful work can be derived from an isolated system
in equilibrium; there must always be an external energy source and a
cold sink.

Do you know what "isolated system" means?

If we isolate an experiment, in a black box, can we lower its
temperature? This is a thermodynamic isolated system. Magnetic fields
are allowed to leave a thermodynamic isolated system.




The second law is often invoked as the reason why perpetual
motion machines cannot exist."

In fact, it has been my experience that most physicists believe in the
above quote. I rest my case!

What case?

No offense, but wake up.



My cooling-water examples are exactly analogous to yours. Do you think
_they_ contradict the 2nd Law?

No answer?

Yes. Again, no offense, but it seems your mind cannot think past 1 ply.
Prior to your example I clarified why certain materials reach
equilibrium at lower temperatures.




Lets simply this for you Richard. Neither experiment A or B have a
power source except thermal noise.

Agreed.

Experiment B radiates more power.

Agreed.

If
you disagree then show the error in the math. It is a very simple
circuit. Over time, more energy is leaving experiment B than experiment
A. Therefore experiment B will be colder than experiment A.

Until it reaches equilibrium.

So?

*!@#$%^&*()_+*!@#$%^&*()_+*!@#$%^&*()_+


So??? The object *was* at room temperature. Without adding any energy
to the system

System. That's an important word in thermodynamics.

What exactly is the "system" here?

The entire experiment is considered the thermodynamically closed
system-- a closed box environment.



Energy is not being added, but is any being taken out of the system?

Yes, in the form of radiation.



What kind of system does your statement of the 2nd Law refer to?
Does _this_ system fit the definition?

I do not adhere to the 2nd law of thermodynamics. I firmly disbelieve
in the illusion of limits.



we successfully lowered the temperature!

Of what? Everything in the isolated system?

Yes. A thermodynamic isolated system does not include everything such
as gravity, magnetism, etc. Such a Law is childish, ridiculous, and
useless. Does your thermodynamic isolated system even prohibit gravity
and magnetism from escaping the system?

***Regardless***, even if _your_ childish and useless rule prohibits
magnetic fields from leaving then the temperature will still lower
because the so-called photons could reflect from one wall to the next.
Therefore, at any given moment there would always be X amount of
so-called photons traversing inside your fictitious universally
isolated system. LOL, you are absolutely hilarious. Although I
seriously doubt that even you Richard would insinuate that a
thermodynamically isolated system pertains to _everything_ included
gravity and magnetism since such a Law is useless and fictitious. How
ridiculous to state, "Hey, if absolutely nothing can leave the system,
.... uh, then gee, guess what, the system still has everything it
started out with. Gosh, now that's darn useful law."



Pull a switch
to disconnect the antenna and now it's not at equilibrium.

It wasn't at equilibrium before. Since energy is leaving the antenna,
you have to consider where it's going to. You can't leave the energy
sink out of an isolated system any more than you can have a battery with
only one terminal.

Are you suggesting magnetic fields cannot leave the antenna in a
thermodynamically isolated system?



You have a
source of so-called "free energy."

Not by any definition a physicist would use.

No offense, but wake up! You are wasting my time.




BTW, my definition of so-called "free energy"

Which one is that? Gibbs free energy? Helmholtz free energy?

merely includes the
process of extracting the energy, not the one time cost of buying the
parts and equipment. Therefore, I term solar cells as a source of
so-called "free energy."

Do I have
to draw you a picture?

Maybe you should draw yourself one. You need to include a heat source, a
heat sink, and a thermal resistance.

If the experiment is in a vacuum then the major factor is black body
radiation.

So add it to your equivalent circuit.

[...]

Most loads, except a
few such as radiation resistance, generate thermal noise.

Anything that turns electrical energy into heat, in fact.
Fluctuation-dissipation theorem.

So Richard Herring believes radiation resistance generates thermal
noise?

Strawman. Since when did radiation resistance turn electrical energy
into *heat* ?

You should be the last person to retort to name-calling. It's not that
becoming of you.

ROFL. You still haven't looked up "strawman", have you?

I sure have. Maybe you haven't. Referring to a person as a Strawman is
an insult by my standards, but it wouldn't surprise me if no word were
an insult by your standards.



Anyhow, you did not answer the question.

That's correct. When you attribute some position to me that I don't
hold, I'm under no obligation to defend it.

You cannot suggest radiation
resistance generates thermal noise simply because electrical resistance
generates thermal noise.

I don't.

If you said, "Well, I detected thermal noise,
therefore radiation resistance must generate thermal noise."

I didn't.

I would
reply, "No, increase the antennas wire thickness and you'll see the
thermal noise decreases, but the radiation resistance remains
relatively the same."

But you can certainly treat the radiation resistance as a noise source,
if the antenna is coupled to other antennas being driven by thermal
sources. Does the name Jansky mean nothing to you?

Evidently not.

That's what I term "2D thinking." You should know better. I would
encourage you to break nature in to its elementary components.
Therefore, do not mix electrical resistance with radiation resistance
and you'll eliminate some of your limitations. It is theoretically
possible to decrease the antennas electrical resistance till it is
appreciably low. Example, the total thermal noise, less antennas
thermal noise, (such as from the power source, etc.) amounts to 1
uV/RtHz, but the antennas thermal noise could be 1 pV/RtHz, or at some
time in the future 1 fV/RtHz by means of future technology.

And what exactly do you think any of that rigmarole has to do with the
antenna's ability to receive radio noise radiated by external thermal
sources?

Receiving the radiation has nothing to do with the experiment. It is of
no concern if the radiation is _absorbed_ by another galaxy or the
Hollywood hills.



Of course, the noise is attenuated by the transmission loss between the
antennas, so its effective temperature is usually very low.

I provided a
broad example by stating "appropriate load." If both the source and
load generate thermal noise then the power across the load is 4 K T dF,
but if the load generates no noise then the power across the load is K
T dF.

I don't know what you think you mean by the power "across" a load. It's
either "into" or "out of".

"Power across a load" means the consumed power as measured across a
load.

Semantic confusion. You measure the voltage "across" the load, but
measure (or infer) the current "through" it, and the sign of their
product is significant.

Do you know a little computer programming? Then you should expand your
mental capacity to a concept called "Indirection."

It's called "indirect measurement" because you have to rely on
additional assumptions.

LOL, assumptions? One could even assume that your entire existence,
Richard, is a dream or some virtual reality in another existence. Have
you watched Matrix the movie?



If those are incorrect there's a more important
programming concept you'll have to reckon with: GIGO.

No offense, but you really appear to be one massively consumed
entangled limitation.



That's why dear
Richard an oscilloscope can display power across a load if you merely
pre-enter the loads resistance. :-)

So if I have a signal generator connected to a 50 ohm resistor via a
coaxial cable, with a T-connector in the middle, and I let you use your
oscilloscope to measure the voltage across it, do you think you can tell
me which way the energy is travelling?

Yes, I can tell you which way the energy is traveling?



Regards,
Paul

.